Heat and Thermodynamics — Page 30 | JEE Physics

Q291

A solid body of constant heat capacity

1

J/{}^ \circ C

is being heated by keeping it in contact with reservoirs in two ways:

(i)

Sequentially keeping in contact with

2

reservoirs such that each reservoir

\,\,\,\,\,\,\,\,

supplies same amount of heat.

(ii)

Sequentially keeping in contact with

8

reservoirs such that each reservoir

\,\,\,\,\,\,\,\,\,\,

supplies same amount of heat. In both the cases body is brought from initial temperature

{100^ \circ }C

to final temperature

{200^ \circ }C

. Entropy change of the body in the two cases respectively is :

A

ln2, 2ln2

B

2ln2, 8ln2

C

ln2, 4ln2

D

ln2, ln2

Correct Answer

Option D

Solution

The entropy change of the body in the two cases is same as entropy is a state function.

Q292

An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity

C

remains constant. If during this process the relation of pressure

P

and volume

V

is given by

P{V^n} =

constant, then

n

is given by (Here

{C_p}

and

{C_v}

are molar specific heat at constant pressure and constant volume, respectively:

A

n = {{{C_p} - C} \over {C - {C_v}}}

B

n = {{C - {C_v}} \over {C - {C_p}}}

C

n = {{{C_p}} \over {{C_v}}}

D

n = {{C - {C_p}} \over {C - {C_v}}}

Correct Answer

Option D

Solution

For a polytropic process

C = {C_v} + {R \over {1 - n}}

$\therefore$

C - {C_v} = {R \over {1 - n}}

$\therefore$

1 - n = {R \over {C - {C_v}}}

$\therefore$

1 - {R \over {C - {C_v}}} = n

$\therefore$

n = {{C - {C_v} - R} \over {C - {C_v}}}

= {{C - {C_v} - {C_p} + {C_v}} \over {C - {C_v}}}

= {{C - {C_p}} \over {C - {C_v}}}

( as

{C_p} - {C_{v = R}}

)

Q293

Match the with {C_v}=\frac{5}{3}$

List - I		List - II
(B)	Diatomic non-rigid gas	(II)	$\dfrac{C_p}{C_v}=\dfrac{7}{5}$
(C)	Monoatomic gas	(III)	$\dfrac{C_p}{C_v}=\dfrac{4}{3}$
(D)	Diatomic rigid gas	(IV)	$\dfrac{C_p}{C_v}=\dfrac{9}{7}$

A A-III, B-IV, C-I, D-II

B A-II, B-IV, C-I, D-III

C A-IV, B-II, C-III, D-I

D A-III, B-II, C-IV, D-I

Correct Answer

Option A

Solution

The explanation uses the equation for the heat capacity ratio, $\gamma = 1 + \dfrac{2}{\text{f}}$ , where $\text{f}$ is the degrees of freedom of the gas.

Here's how it applies to different types of gases: Triatomic Rigid Gas: Degrees of freedom ( $\text{f}$ ) = 6 $\gamma = 1 + \dfrac{2}{6} = \dfrac{4}{3}$ Diatomic Non-Rigid Gas: Degrees of freedom ( $\text{f}$ ) = 7 $\gamma = 1 + \dfrac{2}{7} = \dfrac{9}{7}$ Diatomic Rigid Gas: Degrees of freedom ( $\text{f}$ ) = 5 $\gamma = 1 + \dfrac{2}{5} = \dfrac{7}{5}$ Monoatomic Gas: Degrees of freedom ( $\text{f}$ ) = 3 $\gamma = 1 + \dfrac{2}{3} = \dfrac{5}{3}$ Therefore, when matching List I with List II, the correct pairing is: A (Triatomic rigid gas) - III $(\dfrac{4}{3})$ B (Diatomic non-rigid gas) - IV $(\dfrac{9}{7})$ C (Monoatomic gas) - I $(\dfrac{5}{3})$ D (Diatomic rigid gas) - II $(\dfrac{7}{5})$ Thus, the correct answer from the options given is: A-III, B-IV, C-I, D-II

Q294

Match List - I with List - II. .tg .tg List - I List - II (A) Isobaric (I)

\Delta Q=\Delta W

(B) Isochoric (II)

\Delta Q=\Delta U

(C) Adiabatic (III)

\Delta Q=

zero (D) Isothermal (IV)

\Delta Q=\Delta U+P\Delta V

\Delta Q=

Heat supplied

\Delta W=

Work done by the system

\Delta \mathrm{U}=

Change in internal energy

\mathrm{P}=

Pressure of the system

\Delta \mathrm{V}=

Change in volume of the system Choose the correct answer from the options given below :

A (A)-(IV), (B)-(I), (C)-(III), (D)-(II)

B (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

C (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

D (A)-(II), (B)-(IV), (C)-(III), (D)-(I)

Correct Answer

Option B

Solution

(A) Isobaric $(\mathrm{P}=\mathrm{C})$

\Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{P} \Delta \mathrm{~V}

(B) Isochoric $(\mathrm{V}=\mathrm{C})$

\Delta \mathrm{Q}=\Delta \mathrm{U}

(C) Adiabatic $(\Delta \mathrm{Q}=0)$

\Delta \mathrm{Q}=0

(D) Isothermal $(\Delta \mathrm{U}=0)$

\Delta \mathrm{Q}=\Delta \mathrm{W}

Q295

An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature 300 K. The mean time between two successive collisions is 6

\times

10–8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to

A 0.5

\times

10

-

8 s

B 4

\times

10

-

8 s

C 3

\times

10

-

6 s

D 2

\times

10

-

7 s

Correct Answer

Option B

Solution

t $\propto$

{{Volume} \over {velocity}}

volume $\propto$

{T \over P}

$\therefore$ t $\propto$

{{\sqrt T } \over P}

{{{t_1}} \over {6 \times {{10}^{ - 8}}}} = {{\sqrt {500} } \over {2P}} \times {P \over {\sqrt {300} }}

t1 = 3.8 $\times$ 10 $-$ 8 $\approx$ 4 $\times$ 10 $-$ 8

Q296

A mixture of 2 moles of helium gas (atomic mass = 4 u), and 1 mole of argon gas (atomic mass = 40 u) is kept at 300 K in a container. The ratio of their rms speeds

\left[ {{{{V_{rms}}\,(helium)} \over {{V_{rms}}\,(\arg on)}}} \right],

is close to :

A 3.16

B 0.32

C 0.45

D 2.24

Correct Answer

Option A

Solution

We know, Vrms =

\sqrt {{{3RT} \over M}}

Where M = molar mass of the gas.

Here temperature is 300 K for both the gas.

So temperature is constant.

R is also a constant.

$\therefore$ Vrms

\propto \,\,\sqrt {{1 \over M}}

$\therefore$

{{{V_{rms}}(helium|)} \over {{V_{rms}}\left( {\arg an)} \right)}} = \sqrt {{{{M_{Ar}}} \over {{M_{He}}}}}

=

\sqrt {{{40} \over 4}}

=

\sqrt {10}

= 3.16

Q297

A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27oC. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about : [Take R = 8.3 J/K mole]

A 0.9 kJ

B 6 kJ

C 10 kJ

D 14 kJ

Correct Answer

Option C

Solution

We know, Vrms $\propto$

\sqrt T

So, to make Vrms double we have to make temperature 4 times.

$\therefore$ Final temperature = 300 $\times$ 4 = 1200 K As N2 gas present in the closed vessel So it is a isochoric process.

$\therefore$ Q = nCv

\Delta

T =

{{15} \over {28}} \times \left( {{5 \over 2}R} \right)\left( {1200 - 300} \right)

= 10000 J = 10 kJ

Q298

Match the thermodynamic processes taking place in a system with the correct conditions. In the table :

\Delta

Q is the heat supplied,

\Delta

W is the work done and

\Delta

U is change in internal energy of the system. .tg .tg Process Condition (I) Adiabatic (1)

\Delta

W = 0 (II) Isothermal (2)

\Delta

Q = 0 (III) Isochoric (3)

\Delta

U

\ne

0,

\Delta

W

\ne

0,

\Delta

Q

\ne

0 (IV) Isobaric (4)

\Delta

U = 0

A (I) - (1), (II) - (1), (III) - (2), (IV) - (3)

B (I) - (2), (II) - (4), (III) - (1), (IV) - (3)

C (I) - (1), (II) - (2), (III) - (4), (IV) - (4)

D (I) - (2), (II) - (1), (III) - (4), (IV) - (3)

Correct Answer

Option B

Solution

(I) Adiabatic,

\Delta

Q = 0 (II) Isothermal,

\Delta

U = 0 (III) Isochoric,

\int {pdV}

= 0 $\Rightarrow$ W = 0 (IV) Isobaric process $\Rightarrow$ Pressure remains constant W = P.

\Delta

V $\ne$ 0

\Delta

U $\ne$ 0

\Delta

Q = nCp

\Delta

T $\ne$ 0

Q299

Match the ${{{C_P}} \over {{C_V}}}$ ratio for ideal gases with different type of molecules : .tg .tg Molecule Type CP/CV

List - I		List - II
(A)	Monatomic	(I)	7/5
(B)	Diatomic rigid molecules	(II)	9/7
(C)	Diatomic non-rigid molecules	(III)	4/3
(D)	Triatomic rigid molecules	(IV)	5/3

A (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

B (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

C (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

D (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

Correct Answer

Option C

Solution

\gamma = {C_p}/{C_v}

{\gamma _A} = 1 + {2 \over 3} = 5/3

{\gamma _B} = 1 + {2 \over 5} = 7/5

{\gamma _C} = 1 + {2 \over 7} = 9/7

{\gamma _D} = 1 + {2 \over 6} = 4/3

Q300

Match List I with List II .tg .tg List I List II A. Isothermal Process I. Work done by the gas decreases internal energy B. Adiabatic Process II. No change in internal energy C. Isochoric Process III. The heat absorbed goes partly to increase internal energy and partly to do work D. Isobaric Process IV. No work is done on or by the gas Choose the correct answer from the options given below :

A A-I, B-II, C-IV, D-III

B A-II, B-I, C-III, D-IV

C A-II, B-I, C-IV, D-III

D A-I, B-II, C-III, D-IV

Correct Answer

Option C

Solution

\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}

For isothermal process $\mathrm{T}$ is constant So $\Delta \mathrm{U}=0$ $\mathrm{A} \longrightarrow \mathrm{II}$ Adiabatic process

\begin{aligned} & \Delta \mathrm{Q}=0 \\\\ & \Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W} \\\\ & \Delta \mathrm{U}=-\Delta \mathrm{W} \end{aligned}

Work done by gas is positive So $\Delta \mathrm{U}$ is negative

\text{ B } \longrightarrow \text{ I }

For Isochoric process $\Delta \mathrm{W}=0$

\mathrm{C} \longrightarrow \mathrm{IV}

For Isobaric process

\begin{aligned} & \Delta \mathrm{W}=\mathrm{P} \Delta \mathrm{V} \neq 0 \\\\ & \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T} \neq 0 \end{aligned}

Heat absorbed goes partly to increase internal energy and partly to do work.